Luo Gu -P2258 sub-matrix

Matrices

Title Description

Given the following definition:

1. Matrices: Select a new matrix composed of certain rows and certain columns of a matrix from which the intersection position (relative order to maintain row and column) is called a sub-matrix of the original matrix.

For example, select the second left below 2 2 . 4 4 rows and 2 2 . 4 4 and 5 the element 5 to obtain a crossing position 2 \ 3 Times 2 × FIG. 3 as the right sub-matrix.

9 3 3 3 9

9 4 8 7 4

1 7 4 6 6

6 8 5 6 9

7 4 5 6 1

Wherein a 2 \ 3 Times 2 × submatrix 3 is

4 7 4

8 6 9

2. Adjacent elements: an element of the matrix with its up and down around the four elements (if any) are adjacent.

3. Score matrix: the matrix and the absolute values ​​of the difference of each of the adjacent elements.

The title task: Given a n rows and m columns matrix positive integer, please select a matrix from the r row sub-matrix of column c, such that the minimum value of the sub-matrix, and outputs the value.

Topic links: https://www.luogu.com.cn/problem/P2258

Problem-solving ideas: the rows enumerate (dfs determine which rows selected, the columns dp) can be solved by dp

• #include <bits / STDC ++ H.>
   the using  namespace STD;
   const  int MAXN = 60 ;
   const  int INF = 0x3f3f3f3f ;
   int n-, m, R & lt, C, ANS;
   int R & lt [MAXN], cost [MAXN] [MAXN], DP [MAXN] [MAXN], a [MAXN] [MAXN], Val [MAXN];
   // R & lt need to select the row which represents, cost represents the added value selecting two
   // Val column indicates the increased value selection , dp [i] [j] denotes the front i column select the energy needed j column value 
  int getdp () 
  { 
      int RES = INF;
       for ( int i = . 1 ; i <= m; i ++ ) 
      { 
          Val [i] = 0 ;
          for(int j=1;j<r;j++)
          {
              val[i]+=abs(a[R[j]][i]-a[R[j+1]][i]);
          }
      }
      for(int i=1;i<=m;i++)
      {
          for(int j=i+1;j<=m;j++)
          {
              cost[i][j]=0;
              for(int k=1;k<=r;k++)
              {
                  cost[i][j]+=abs(a[R[k]][i]-A [R & lt [K]] [J]); 
              } 
          } 
      } 
      for ( int i = . 1 ; i <= m; i ++) // assumed that the i-th column has been selected, only dp i-1 j- selected column before a required minimum value to 
      {
           for ( int J = . 1 ; J <= I && J <= C; J ++ ) 
          { 
              DP [I] [J] = 1E9;
               for ( int K = J- . 1 ; K <I ; K ++ ) 
                DP [I] [J] = min (DP [I] [J], DP [K] [J- . 1 ] + cost [K] [I] + Val [I]); 
          } 
      } 
      for ( int i=c;i<=m;i++) res=min(res,dp[i][c]);
      return res;
  }
  void dfs(int now,int cnt)
  {
      if(now>n)
      {
          if(cnt==r) ans=min(ans,getdp());
          return ;
      }
      dfs(now+1,cnt);
      R[cnt+1]=now;
      dfs(now+1,cnt+1);
      return ;
  }
  int main()
  {
      cin>>n>>m>>r>>c;
       for(int i=1;i<=n;i++)
       {
           for(int j=1;j<=m;j++)
           {
                cin>>a[i][j];
           }
  
       }
       ans=inf;
       dfs(1,0);
       cout<<ans<<endl;
       return 0;
  }